Question and Answers Forum

ArithmeticQuestion and Answers: Page 1

Question Number 222736 Answers: 0 Comments: 0

Question Number 222352 Answers: 1 Comments: 0

Prove that : (a−b)(a−c)(a−d)(b−c)(b−d)(c−d) divisible by 12, with a,b,c,d ∈Z

$${Prove}\:{that}\::\:\left({a}−{b}\right)\left({a}−{c}\right)\left({a}−{d}\right)\left({b}−{c}\right)\left({b}−{d}\right)\left({c}−{d}\right)\:{divisible}\:{by}\:\mathrm{12},\:{with}\:{a},{b},{c},{d}\:\in\mathbb{Z} \\ $$

Question Number 222249 Answers: 3 Comments: 0

Prove:∀n∈Z^+ ,1^3 +2^3 +…+n^3 =(1+2+…+n)^2

$$\mathrm{Prove}:\forall{n}\in\mathbb{Z}^{+} ,\mathrm{1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +\ldots+{n}^{\mathrm{3}} =\left(\mathrm{1}+\mathrm{2}+\ldots+{n}\right)^{\mathrm{2}} \\ $$

Question Number 222097 Answers: 0 Comments: 2

Question Number 221787 Answers: 1 Comments: 0

(√((1−4x(√(1−4x^2 )))/2)) = 1−8x^2 x=?

$$\:\:\sqrt{\frac{\mathrm{1}−\mathrm{4x}\sqrt{\mathrm{1}−\mathrm{4x}^{\mathrm{2}} }}{\mathrm{2}}}\:=\:\mathrm{1}−\mathrm{8x}^{\mathrm{2}} \\ $$$$\:\:\mathrm{x}=?\: \\ $$

Question Number 221247 Answers: 1 Comments: 1

Question Number 220738 Answers: 1 Comments: 0

Question Number 220737 Answers: 2 Comments: 0

Question Number 220160 Answers: 0 Comments: 0

for all x , y [0 , 1] ; prove that; [ (((x^3 + y^3 + 𝛇(3)))^(1/(3 )) /(1 + e^(−x^2 y^2 ) )) + (((x^4 + 𝚪(y+1)))^(1/(4 )) /((1 + y^2 )^(1/3) )) + ((ln(1 + x^5 + y^5 ))/( (√(1 + x^2 + y^2 )))) + Li_2 (xy) + ((√(x^6 + y^6 +1 ))/((1 + x^3 y^3 )^(1/2) )) ≤ (e^(xy) /(1 + x + y )) + ((ln (1 + x^2 + y^2 ) ))^(1/(3 )) + ((2𝛇(2))/( (√(1 + x^2 y^2 )))) + ((x^8 + y^8 + 1))^(1/(4 )) ]

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{for}\:\mathrm{all}\:{x}\:,\:{y}\:\left[\mathrm{0}\:,\:\mathrm{1}\right]\:;\:\mathrm{prove}\:\mathrm{that}; \\ $$$$\:\:\left[\:\frac{\sqrt[{\mathrm{3}\:\:}]{\boldsymbol{{x}}^{\mathrm{3}} \:+\:\boldsymbol{{y}}^{\mathrm{3}} \:+\:\boldsymbol{\zeta}\left(\mathrm{3}\right)}}{\mathrm{1}\:+\:\boldsymbol{{e}}^{−\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{y}}^{\mathrm{2}} } \:}\:+\:\frac{\sqrt[{\mathrm{4}\:\:}]{\boldsymbol{{x}}^{\mathrm{4}} +\:\boldsymbol{\Gamma}\left(\boldsymbol{{y}}+\mathrm{1}\right)}}{\left(\mathrm{1}\:+\:\boldsymbol{{y}}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{3}} }\:+\:\frac{\boldsymbol{\mathrm{ln}}\left(\mathrm{1}\:+\:\boldsymbol{{x}}^{\mathrm{5}} \:+\:\boldsymbol{{y}}^{\mathrm{5}} \right)}{\:\sqrt{\mathrm{1}\:+\:\boldsymbol{{x}}^{\mathrm{2}} \:+\:\boldsymbol{{y}}^{\mathrm{2}} }}\:\:\:\:\:\:\:\:\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\boldsymbol{\mathrm{Li}}_{\mathrm{2}} \left(\boldsymbol{{xy}}\right)\:+\:\frac{\sqrt{\boldsymbol{{x}}^{\mathrm{6}} \:+\:\boldsymbol{{y}}^{\mathrm{6}} \:+\mathrm{1}\:}}{\left(\mathrm{1}\:+\:\boldsymbol{{x}}^{\mathrm{3}} \boldsymbol{{y}}^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{2}} }\: \\ $$$$\left.\:\:\:\:\:\:\leqslant\:\frac{\boldsymbol{{e}}^{\boldsymbol{{xy}}} }{\mathrm{1}\:+\:\boldsymbol{{x}}\:+\:\boldsymbol{{y}}\:}\:+\:\sqrt[{\mathrm{3}\:\:}]{\boldsymbol{\mathrm{ln}}\:\left(\mathrm{1}\:+\:\boldsymbol{{x}}^{\mathrm{2}} \:+\:\boldsymbol{{y}}^{\mathrm{2}} \right)\:}\:+\:\frac{\mathrm{2}\boldsymbol{\zeta}\left(\mathrm{2}\right)}{\:\sqrt{\mathrm{1}\:+\:\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{y}}^{\mathrm{2}} }}\:+\:\sqrt[{\mathrm{4}\:\:}]{\boldsymbol{{x}}^{\mathrm{8}} \:+\:\boldsymbol{{y}}^{\mathrm{8}} \:+\:\mathrm{1}}\:\:\:\:\:\:\:\:\right]\:\: \\ $$$$ \\ $$$$\:\: \\ $$

Question Number 220159 Answers: 0 Comments: 1

for all x, y ∈ [0 , 1] ; prove that; (1/( (√(1 + x^4 )))) + (2/( (√(1 + y^4 )))) + (2/( (√(4 + (x + y)^4 )))) + ((2(√2))/( (√(2+ x^2 y^2 + y^3 )))) ≤ (2/( (√(1 + x^2 y^2 )))) + (2/(^4 (√(1 + x^5 + y^5 )))) + ln(e+((x^3 y+y^3 x)/(1 + xy))) + (1/((1+x+y)^3 ))

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{for}\:\mathrm{all}\:{x},\:{y}\:\in\:\left[\mathrm{0}\:,\:\mathrm{1}\right]\:;\:\mathrm{prove}\:\mathrm{that}; \\ $$$$\:\:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{4}} }}\:+\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}\:+\:{y}^{\mathrm{4}} }}\:+\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{4}\:+\:\left({x}\:+\:{y}\right)^{\mathrm{4}} }}\:+\:\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}+\:{x}^{\mathrm{2}} {y}^{\mathrm{2}} \:+\:{y}^{\mathrm{3}} }}\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\leqslant\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{2}} {y}^{\mathrm{2}} }}\:+\:\frac{\mathrm{2}}{\:^{\mathrm{4}} \sqrt{\mathrm{1}\:+\:{x}^{\mathrm{5}} \:+\:{y}^{\mathrm{5}} }}\:+\:\mathrm{ln}\left({e}+\frac{{x}^{\mathrm{3}} {y}+{y}^{\mathrm{3}} {x}}{\mathrm{1}\:+\:{xy}}\right)\:+\:\frac{\mathrm{1}}{\left(\mathrm{1}+{x}+{y}\right)^{\mathrm{3}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 220096 Answers: 1 Comments: 0

let a, b, c, d, e is a positive real numbers and K = a + b + c + d + e +1 . prove that; Σ_(cyc) (1/(k−a)) < (1/4) ((((e^3 d^3 c))^(1/(4 )) /(c^(3/4) d^(1/2) e^(1/4) (√a))) + (((d^( 3) c^2 b))^(1/(4 )) /(d^( 3/4) c^(1/2) b^(1/4) (√e))) + (((c^3 b^2 a))^(1/(4 )) /(c^(3/4) b^(1/2) a^(1/4) (√d))) + (((b^3 a^2 e))^(1/(4 )) /(b^(3/4) a^(1/2) e^(1/4) (√c))) + (((a^3 e^2 d))^(1/(4 )) /(a^(3/4) e^(1/2) d^(1/4) (√b))))

$$ \\ $$$$\:\:\:\mathrm{let}\:{a},\:{b},\:{c},\:{d},\:{e}\:\mathrm{is}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{and} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{K}\:=\:{a}\:+\:{b}\:+\:{c}\:+\:{d}\:+\:{e}\:+\mathrm{1}\:. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{prove}\:\mathrm{that}; \\ $$$$\:\:\:\:\underset{\boldsymbol{{cyc}}} {\sum}\:\frac{\mathrm{1}}{\boldsymbol{{k}}−\boldsymbol{{a}}}\:<\:\frac{\mathrm{1}}{\mathrm{4}}\:\left(\frac{\sqrt[{\mathrm{4}\:\:\:\:}]{\boldsymbol{{e}}^{\mathrm{3}} \boldsymbol{{d}}^{\mathrm{3}} \boldsymbol{{c}}}}{\boldsymbol{{c}}^{\mathrm{3}/\mathrm{4}} \boldsymbol{{d}}^{\mathrm{1}/\mathrm{2}} \boldsymbol{{e}}^{\mathrm{1}/\mathrm{4}} \sqrt{\boldsymbol{{a}}}}\:+\:\frac{\sqrt[{\mathrm{4}\:\:\:}]{\boldsymbol{{d}}^{\:\mathrm{3}} \boldsymbol{{c}}^{\mathrm{2}} \boldsymbol{{b}}}}{\boldsymbol{{d}}^{\:\mathrm{3}/\mathrm{4}} \boldsymbol{{c}}^{\mathrm{1}/\mathrm{2}} \boldsymbol{{b}}^{\mathrm{1}/\mathrm{4}} \sqrt{\boldsymbol{{e}}}}\:+\:\frac{\sqrt[{\mathrm{4}\:\:}]{\boldsymbol{{c}}^{\mathrm{3}} \boldsymbol{{b}}^{\mathrm{2}} \boldsymbol{{a}}}}{\boldsymbol{{c}}^{\mathrm{3}/\mathrm{4}} \boldsymbol{{b}}^{\mathrm{1}/\mathrm{2}} \boldsymbol{{a}}^{\mathrm{1}/\mathrm{4}} \sqrt{\boldsymbol{{d}}}}\:+\:\frac{\sqrt[{\mathrm{4}\:\:}]{\boldsymbol{{b}}^{\mathrm{3}} \boldsymbol{{a}}^{\mathrm{2}} \boldsymbol{{e}}}}{\boldsymbol{{b}}^{\mathrm{3}/\mathrm{4}} \boldsymbol{{a}}^{\mathrm{1}/\mathrm{2}} \boldsymbol{{e}}^{\mathrm{1}/\mathrm{4}} \sqrt{\boldsymbol{{c}}}}\:+\:\frac{\sqrt[{\mathrm{4}\:\:}]{\boldsymbol{{a}}^{\mathrm{3}} \boldsymbol{{e}}^{\mathrm{2}} \boldsymbol{{d}}}}{\boldsymbol{{a}}^{\mathrm{3}/\mathrm{4}} \boldsymbol{{e}}^{\mathrm{1}/\mathrm{2}} \boldsymbol{{d}}^{\mathrm{1}/\mathrm{4}} \sqrt{\boldsymbol{{b}}}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$\:\: \\ $$

Question Number 219933 Answers: 1 Comments: 0

A^1 + B^2 + C^3 + D^4 = ABCD^(−) find ABCD

$$\mathrm{A}^{\mathrm{1}} \:+\:\mathrm{B}^{\mathrm{2}} \:+\:\mathrm{C}^{\mathrm{3}} \:+\:\mathrm{D}^{\mathrm{4}} \:=\:\overline {\mathrm{ABCD}} \\ $$$${find}\:\:{ABCD} \\ $$

Question Number 219874 Answers: 2 Comments: 0

find (√2^6^2^1^4^4 )=?

$${find}\:\sqrt{\mathrm{2}^{\mathrm{6}^{\mathrm{2}^{\mathrm{1}^{\mathrm{4}^{\mathrm{4}} } } } } }=? \\ $$

Question Number 219556 Answers: 0 Comments: 0

Question Number 219085 Answers: 0 Comments: 0

Question Number 219084 Answers: 0 Comments: 0

Question Number 219083 Answers: 0 Comments: 0

Question Number 218889 Answers: 5 Comments: 0

Question Number 218834 Answers: 0 Comments: 0

Question Number 218832 Answers: 0 Comments: 0

Question Number 218799 Answers: 0 Comments: 1

For those who are interested in cryptography. The below text has been encrypted using Vigenere cipher, such that numbers, punctuation marks and the letter E^(..) have remained the same. A keyword of length 9 has been used, which starts with the letter K. Decrypt the text.

$$\mathrm{For}\:\mathrm{those}\:\mathrm{who}\:\mathrm{are}\:\mathrm{interested}\:\mathrm{in}\:\mathrm{cryptography}. \\ $$$$\mathrm{The}\:\mathrm{below}\:\mathrm{text}\:\mathrm{has}\:\mathrm{been}\:\mathrm{encrypted}\:\mathrm{using} \\ $$$$\mathrm{Vigenere}\:\mathrm{cipher},\:\mathrm{such}\:\mathrm{that}\:\mathrm{numbers},\:\mathrm{punctuation} \\ $$$$\mathrm{marks}\:\mathrm{and}\:\mathrm{the}\:\mathrm{letter}\:\overset{..} {\mathrm{E}}\:\mathrm{have}\:\mathrm{remained}\:\mathrm{the}\:\mathrm{same}. \\ $$$$\mathrm{A}\:\mathrm{keyword}\:\mathrm{of}\:\mathrm{length}\:\mathrm{9}\:\mathrm{has}\:\mathrm{been}\:\mathrm{used},\:\mathrm{which} \\ $$$$\mathrm{starts}\:\mathrm{with}\:\mathrm{the}\:\mathrm{letter}\:\mathrm{K}.\:\mathrm{Decrypt}\:\mathrm{the}\:\mathrm{text}. \\ $$

Question Number 218675 Answers: 4 Comments: 0

Question Number 218312 Answers: 1 Comments: 0

Question Number 218066 Answers: 1 Comments: 2

If Alphaprime borrows 25$ from LYCONTRIX at the rzte of 12% p.a. How much does the former owe to the other in a span of 8years?

$${If}\:{Alphaprime}\:{borrows}\:\mathrm{25\$}\:{from} \\ $$$${LYCONTRIX}\:{at}\:{the}\:{rzte}\:{of}\:\mathrm{12\%}\:{p}.{a}. \\ $$$${How}\:{much}\:{does}\:{the}\:{former}\:{owe}\:{to}\:{the} \\ $$$${other}\:{in}\:{a}\:{span}\:{of}\:\mathrm{8}{years}? \\ $$

Question Number 217278 Answers: 0 Comments: 0

Question Number 216726 Answers: 0 Comments: 4

Pg 1 Pg 2 Pg 3 Pg 4 Pg 5 Pg 6 Pg 7 Pg 8 Pg 9 Pg 10

Contact: info@tinkutara.com